Rule A5-1-7 (required, implementation, automated)
A lambda shall not be an operand to decltype or typeid.
Rationale
“The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed non-union class type [...]” [C++14 Language Standard] [3] Each lambda expression has a different unique underlying type, and therefore the type is not to be used as a decltype or typeid argument. It is allowed to use it as a template parameter and a function argument.
Example
// $Id: A5-1-7.cpp 289815 2017-10-06 11:19:11Z michal.szczepankiewicz $
#include <cstdint>
#include <functional>
#include <vector>
void Fn()
{
auto lambda1 = []() -> std::int8_t { return 1; };
auto lambda2 = []() -> std::int8_t { return 1; };
if (typeid(lambda1) == typeid(lambda2)) // Non-compliant - types of lambda1
// and lambda2 are different
{
// ...
}
std::vector<decltype(lambda1)> v; // Non-compliant
// v.push_back([]() { return 1; }); // Compilation error, type of pushed
// lambda is different than decltype(lambda1)
// using mylambda_t = decltype([](){ return 1; }); // Non-compliant // compilation error
auto lambda3 = []() { return 2; };
using lambda3_t = decltype(lambda3); // Non-compliant - lambda3_t type can
// not be used for lambda expression
// declarations
// lambda3_t lambda4 = []() { return 2; }; // Conversion error at
// compile-time
std::function<std::int32_t()> f1 = []() { return 3; };
std::function<std::int32_t()> f2 = []() { return 3; };
if (typeid(f1) == typeid(f2)) // Compliant - types are equal
{
// ...
}
}
template <typename T>
void Foo(T t);
void Bar()
{
Foo([]() {}); // Compliant
}
See also
none